Most tests often include questions based on the knowledge of the geometries of 3-D objects such as cylinder, cone, cuboid, cube & sphere. The purpose of the

article is to help you learn basics of 3-D geometry and encapsulate some of the important formulae and tricks.

The questions on Volume and Surface Area appear in all the competitive exams. Most of the students tend to avoid this topic considering it to be quite complex and calculative. This article would help you not only in memorizing the formulas, but also in understanding direct or indirect applications of these formulas. We strongly advice you go through each and every definition and formula given below to solve questions on Surface Area and Volume.

Surface area and Volume Formulas

Solids: Solids are three–dimensional objects, bound by one or more surfaces. Plane surfaces of a solid are called its faces. The lines of intersection of adjacent faces are called edges.

For any regular solid, Number of faces + Number of vertices = Number of edges + 2. This formula is called Euler’s formula.

Volume: Volume of a solid figure is the amount of space enclosed by its bounding surfaces. Volume is measured in cubic units.

Sphere: The set of all points in space, which are at a fixed distance from a fixed point, is called a Sphere. The fixed point is the centre of the sphere and the fixed distance is the radius of the sphere.

Volume = 4/3πr^{3}. Surface Area (curved and total) = 4 πr^{2}.

1. The radius of sphere is increased by 40 %. The increase in surface area of the sphere is:

1. 100%

2. 125%

3. 96%

4. 160%

Explanation Radius of sphere increased by 40 %. ∴ new surface area = 1.4 x 1.4 ⇒ 1.96
% increase in surface area =(1.96 - 1 / 1) x 100 = .96 x 100 = 96%

2. A copper sphere of diameter 12 cm is drawn into a wire of diameter 8 mm. Find the length of the wire.

1. 19,440 cm

2. 1,800 cm

3. 30,520 cm

4. 48,600 cm

Explanation: Here 4/3 πR^{3} = πr^{2}h where R is the radius of sphere, r is the radius of the wire and h is the length of the wire.
Hence 4/3 (12/2)^{3} = (.8/2)^{2} h
So h =1800cm.

Hemisphere: A sphere cut by a plane passing through its centre forms two hemispheres. The upper surface of a hemisphere is a circular region.
Volume = 2/3πr^{3}. Surface Area (curved) =2πr^{2}. Surface Area (Total) = 2πr^{2} + πr^{2} (formula for area- plane) ⇒ 3πr^{2}.

3. A cylinder and a hemisphere have equal volumes and equal bases. Find the ratio of height of cylinder to height of hemisphere.

1. 2:1

2. 3:1

3. 4:1

4. 5:1

5. 1:2

Explanation: Volume of cylinder = πr^{2}h
Volume of hemisphere = 2/3 πr^{3}
Equating the two you get the ratio as h/r = 4/3.
This is basically the ratio their heights only.
Hence the answer is 4 : 3.

4. In the toy shown in the adjoining figure, the height of the cone = 15 cm, total height of the toy = 40 cm, radius of the hemisphere = 7 cm. Find the volume of the toy. (π = 22/7)

1. 1356.69 cm^{3}

2. 4260.67 cmsup>3

3. 1065.16 cmsup>3

4. None of these

Explanation:
Volume of toy = volume of cone + volume of cylinder + volume of hemisphere
= (1/3 x π x 7^{2} x 15) +{π x 7^{2} x 18} + [2/3 x π x 7^{3}]
= 49π (5 + 18 + 14/3) = 4260.67 cm^{3}.

Spherical shell: If R and r are the outer and inner radius of a hollow sphere, then volume of material in a spherical shell = 4/3π (R^{3} – r^{3}).

5. A solid hemisphere of radius 'a' that is made of a certain material weighs 60 kg. The figure is given below. What is the weight, in kg, of the entire hemispherical shell if the outer radius is '7a' and the inner radius is 3a?

1. 19888

2. 19860

3. 16890

4. 1,9080

5. 18960

Explanation: 5
V = 2/3πa^{3} and this volume weighs 60 kgs.
If it is a shell, you can find V of solid part by subtracting inner empty hemisphere from total.
V of entire sphere 2/3π(7a)^{3} and V of inner empty space = 2/3π(3a)^{3}
Therefore, V of solid = 2/3π(7a)^{3} - 2/3π(3a)^{3} = 2/3πa^{3} (343 - 27) = 2/3πa^{3} x 316
Now 2/3πa^{3} weighs 60 kg,
so, 2/3πa^{3} x 316 will weigh 60 x 316=18960kg.